If a, b, c are sides opposte to the angles A, B, C then which of the following is correct (1)`(b+c) cos (A/2)=a sin ((B+C)/2)` (2) `(b+c)cos ((B+C)/2)= a sin (A/2)`(3) `(b-c) cos((B-C)/2)= a cos A/2` (4) `(b-c) cos (A/2)=a sin ((B-C)/2)`

In triangle ABC,a,b,c are the lengths of its sides and A,B,C are the angles of triangle ABC .The correct relation is given by (a) (b-c)sin((B-C)/(2))=a(cos A)/(2) (b) (b-c)cos((A)/(2))=as in(B-C)/(2)(c)(b+c)sin((B+C)/(2))=a(cos A)/(2)(d)(b-c)cos((A)/(2))=2a(sin(B+C))/(2)

If A, B, C are interior angle of triangle ABC then show that sin ((A+B)/2) + cos (( A+ B)/2) = cos (C/2) + sin (C/2)

a cos A + b cos B + c cos C = 2a sin B sin C

(cos ^(2) ((B-C)/( 2)) )/( (b+c)^(2))+( sin ^(2)((B-C)/( 2)) )/( (b-c)^(2))=

a.cos A + b.cos B + c.cos C = 2a sin B sin C

cos A + cos B + cos C = 1 + 4sin ((A) / (2)) sin ((B) / (2)) sin ((C) / (2))

If A+B+C=pi , prove that : sin ((B+C)/(2)) + sin ((C+A)/(2)) + sin( (A+B)/(2) )= 4cos ((pi-A)/(4)) cos( (pi-B)/(4)) cos((pi-C)/(4)) .

sin A + sin B + sin C = 4cos ((A) / (2)) cos ((B) / (2)) cos ((C) / (2))